\newproblem{lay:5_4_22}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 5.4.22}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	If $A$ is diagonalizable and $B$ is similar to $A$, then $B$ is also diagonalizable.
}{
  % Solution
	If $A$ is diagonalizable, there exist an invertible matrix $P$ and a diagonal matrix $D$ such that
	\begin{center}
		$A=PDP^{-1}$
	\end{center}
	If $B$ is similar to $A$, then there exists an invertible matrix $Q$ such that
	\begin{center}
		$B=QAQ^{-1}$
	\end{center}
	Combining both results we have
	\begin{center}
		$B=Q(PDP^{-1})Q^{-1}=(QP)D(P^{-1}Q^{-1})$
	\end{center}
	So $B$ is also diagonalizable since it can be expressed as
	\begin{center}
		$B=P'D(P')^{-1}$
	\end{center}
	being $P'=QP$ an invertible matrix and $D$ a diagonal matrix.
}
\useproblem{lay:5_4_22}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
